POJ 2001:Shortest Prefixes
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 16782 | Accepted: 7286 |
Description
is considered to be a prefix of itself. In everyday language, we tend to abbreviate words by prefixes. For example, "carbohydrate" is commonly abbreviated by "carb". In this problem, given a set of words, you will find for each word the shortest prefix that
uniquely identifies the word it represents.
In the sample input below, "carbohydrate" can be abbreviated to "carboh", but it cannot be abbreviated to "carbo" (or anything shorter) because there are other words in the list that begin with "carbo".
An exact match will override a prefix match. For example, the prefix "car" matches the given word "car" exactly. Therefore, it is understood without ambiguity that "car" is an abbreviation for "car" , not for "carriage" or any of the other words in the list
that begins with "car".
Input
Output
Sample Input
carbohydrate
cart
carburetor
caramel
caribou
carbonic
cartilage
carbon
carriage
carton
car
carbonate
Sample Output
carbohydrate carboh
cart cart
carburetor carbu
caramel cara
caribou cari
carbonic carboni
cartilage carti
carbon carbon
carriage carr
carton carto
car car
carbonate carbona
题意:给出若干字符串,求把他们缩写所能形成的最小程度。
如:
abdef
abcdefg
ab
其缩写最小程度为
abd
abc
ab
可以使用字典树求出衍生边为1的各个单词,然后输出!
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct tree
{
char c;
struct tree *next[27];
int num;
};
void init(char *c,tree *T)
{
tree *p;
for(int i=0; i<(int)strlen(c); i++)
{
if(T->next[c[i]-'a']==NULL)
{
p=(tree*)malloc(sizeof(tree));
T->next[c[i]-'a']=p;
p->c=c[i];
T=T->next[c[i]-'a'];
memset(T->next,0,sizeof(T->next));
T->num=1;
}
else
{
T=T->next[c[i]-'a'];
T->num++;
}
}
}
void PR(char *c,tree *t)
{
t=t->next[c[0]-'a'];
for(int i=0; i<(int)strlen(c); i++)
{
if(t->num>1)
{
printf("%c",t->c);
t=t->next[c[i+1]-'a'];
}
else
{
printf("%c",t->c);
break;
}
}
}
int main()
{
char c[1005][35]={{0}};
int i;
tree *T=(tree*)malloc(sizeof(tree));
T->num=0;
memset(T->next,0,sizeof(T->next));
for(i=0; gets(c[i]); i++)
{
init(c[i],T);
}
for(int k=0; k<i; k++)
{
printf("%s ",c[k]);
PR(c[k],T);
printf("\n");
}
return 0;
}
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